See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material. The starting material is 4-(2-hydroxypropyl)benzaldehyde: a benzene ring bearing a –CHO group on one end and a –CH2–CH(OH)–CH3 group on the para position. Step 2 – Reaction with excess EtMgBr (Grignard reagent). The Grignard reagent (CH3CH2MgBr) is present in excess, so it reacts with every electrophilic carbonyl/OH-derived site. First, the aldehyde (–CHO) reacts with one equivalent of EtMgBr to give, after acidic workup (H+), a secondary alcohol: –CH(OH)–CH2CH3 at the position formerly bearing the aldehyde. The secondary alcohol already present in the side chain (–CH(OH)–CH3) does not react directly with Grignard under normal conditions (the OH proton is deprotonated but does not add). So after H+ workup, the product has two OH groups: one newly formed secondary alcohol from the aldehyde addition (–CH(OH)–Et at the ring) and the original –CH(OH)–CH3 in the side chain. Step 3 – Treatment with H+ and heat (Delta): acid-catalyzed dehydration. Under acidic conditions with heating, secondary alcohols undergo dehydration (E1 or E2) to form alkenes. Both OH groups can dehydrate. - The newly formed alcohol –Ar–CH(OH)–CH2CH3 dehydrates to give Ar–CH=CH–CH3 (an internal alkene conjugated with the ring, a styrene-type system). This alkene has E and Z isomers → 2 geometrical isomers. - The side chain alcohol –CH2–CH(OH)–CH3 dehydrates to give –CH2–CH=CH2 (terminal alkene, no geometrical isomers) or –CH=CH–CH3 (internal alkene, E/Z possible) depending on which beta-H is removed. - If dehydration gives –CH=CH–CH3 from the benzylic position, this is also an internal alkene with E and Z isomers → 2 geometrical isomers. - If dehydration gives –CH2–CH=CH2, there is only one form (terminal, no E/Z). Step 4 – Count geometrical isomers of (X). The product (X) results from both dehydrations. The two alkene-forming positions each can give E or Z: the ring-side alkene (2 isomers: E and Z) combined with the side-chain alkene giving –CH=CH–CH3 (2 isomers: E and Z) gives 2 × 2 = 4 geometrical isomers total. Step 5 – Verify. With two independent double bonds each capable of E/Z isomerism, the total number of geometrical isomers = 2 × 2 = 4. This matches the given answer. Therefore, the correct answer is 4.